Question

A, B, C and D are points in a vertical line such that AB = BC = CD. If a body falls from rest from A, then the times of descend through AB, BC and CD are in ratio:-

• A. $1:\sqrt{2}:\sqrt{3}$
• B. $\sqrt{2}:\sqrt{3}$ : 1
• C. $\sqrt{3}:1:\sqrt{2}$
• D. $1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Answer 1

Answer: (D)

$1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

In above figure,

A body released $\Rightarrow u=0$

Let $h_1=\frac{1}{2}g{t}_1^2$

$\Rightarrow t_1=\sqrt{\frac{2h}{g}}\times 1...\left(1\right)$

$\Rightarrow t_1+t_2=\sqrt{\frac{2\left(2h\right)}{g}}$

$\Rightarrow t_2=\sqrt{\frac{2h}{g}}\times (\sqrt{2}-1)...\left(2\right)$

Now, $t_1+t_2+t_3=\sqrt{\frac{2\left(3h\right)}{g}}$

$\Rightarrow t_3=\sqrt{\frac{6h}{g}}-\sqrt{\frac{4h}{g}}$

$\Rightarrow t_3=\sqrt{\frac{2h}{g}}\times (\sqrt{3}-\sqrt{2})...\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

$t_1:t_2:t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Hence, $\text{D}$ is correct option.